给定一个可以是正数和负数的数字序列,有几种算法可以找到最长的增长子序列。但是,如果存在多个最长的增加子序列,有人可以给我一个算法来找到具有最大总和的最长增长子序列吗?
示例:对于20,1,4,3,10,答案是1,4,10,而不是1,3,10
答案 0 :(得分:3)
dpLen[i] = maximum length of a LIS with maximum sum ending at i
dpSum[i] = maximum sum of a LIS with maximum sum ending at i
for i = 0 to n do
dpLen[i] = 1
dpSum[i] = input[i]
maxLen = 0
for j = 0 to i do
if dpLen[j] > maxLen and input[j] < input[i]
maxLen = dpLen[j]
for j = 0 to i do
if dpLen[j] == maxLen and input[j] < input[i] and dpSum[j] + input[i] > dpSum[i]
dpSum[i] = dpSum[j] + input[i]
dpLen[i] = maxLen + 1
答案 1 :(得分:2)
这是一个动态编程问题。这是一个有效的例子。我试图注释代码。但是,如果你最近没有弄清楚动态编程概念,那么很难理解解决方案。
解决方案可以被认为是
S(j)= max { 以j结尾的最长和子序列的和(即包括[j]), S(J-1) }
public class LongestSumSequence{
public static void printLongestSumSubsequence(int[] seq) {
int[] S = new int[seq.length];
//S[j] contains the longest sum of subsequence a1,a2,a3,....,aj
//So a sub sequence with length 1 will only contain first element.
//Hence we initialize it like this
S[0] = seq[0];
int min_index = 0;
int max_index = 0;
//Now like any dynamic problem we proceed by solving sub problems and
//using results of subproblems to calculate bigger problems
for(int i = 1; i < seq.length; i++) {
//Finding longest sum sub-sequence ending at j
int max = seq[i];
int idx = i;
int sum = seq[i];
for(int j = i-1; j >=0 ; j--) {
sum += seq[j];
if(max < sum) {
idx = j;
max = sum;
}
}
//Now we know the longest sum subsequence ending at j, lets see if its
//sum is bigger than S(i-1) or less
//This element is part of longest sum subsequence
if(max > S[i-1]) {
S[i] = max;
max_index = i;
min_index = idx;
} else {
//This element is not part of longest sum subsequence
S[i] = S[i-1];
}
}
System.out.println("Biggest Sum : "+S[seq.length - 1]);
//Print the sequence
for(int idx = min_index; idx <= max_index; idx++) {
System.out.println("Index " + idx + "Element " + seq[idx]);
}
}
public static void main(String[] args) {
int[] seq = {5,15,-30,10,-5,40,10};
printLongestSumSubsequence(seq);
}
}